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Another rather simple Differential Equation

  1. Jun 1, 2006 #1
    [tex]y \prime = (cos^{2}x)(cos^{2}(2y))[/tex]

    [tex] \frac{dy}{dx} = \frac{1}{2} (cox(2x) + 1) \frac{1}{4}(cos(4y) + 1)[/tex]

    [tex] \frac{1}{cos(4y) + 1}dy \ = \ \frac{1}{8} (cos(2x) + 1)dx [/tex]

    [tex] \int \frac{1}{ \sqrt{cos(4y)}^{2} + 1}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx [/tex]

    [tex] tan^{-1}(cos4y) \ = \ \frac{1}{8} \frac{1}{2} sin(2x) + \frac{1}{8} x + C [/tex]

    [tex] tan[tan^{-1}(cos4y) \ = \ \frac{1}{16} sin(2x) + \frac{1}{8} x + C] [/tex]

    [tex] cos^{-1}[cos(4y) \ = \ tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]

    [tex] y \ = \ \frac{1}{4} cos^{-1}[tan( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]
     
    Last edited: Jun 1, 2006
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  3. Jun 1, 2006 #2

    Hootenanny

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    What's the question? What have you attempted?

    ~H
     
  4. Jun 1, 2006 #3
    The first line is the problem, and each following line is one of the steps I took to solve it. The final line is the solution. Is that right? It sure doesn't look right to me, but I don't know what I am doing wrong.
     
    Last edited: Jun 1, 2006
  5. Jun 1, 2006 #4

    Hootenanny

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    I am a little concerned as to what you have done with respect to this line here;

    [tex] \int \frac{1}{{\color{red} \sqrt{cos(4y)}^{2} + 1}}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx [/tex]

    Why did you change it into this form? Why did you simply not integrate;

    [tex]\int \frac{1}{cos(4y) + 1}dy[/tex]

    In any case

    [tex] \frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}[/tex]

    ~H
     
  6. Jun 1, 2006 #5
    Why is this so:

    [tex] \frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}[/tex]

    And I changed it into that form because I thought that would be an easy way to use the arctangent integration technique.

    How would I integrate this:
    [tex]\int \frac{1}{cos(4y) + 1}dy[/tex]
     
    Last edited: Jun 1, 2006
  7. Jun 1, 2006 #6
    Oh, would this work:

    [tex]\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})[/tex]

    I reworked the problem that way, and this is what I ended up with:

    [tex] y \ = \ \frac{1}{4} cos^{-1}[tan^{2}( \frac{sin2x}{16} + \frac{x}{8} + C)] [/tex]

    Am I getting warmer?
     
    Last edited: Jun 1, 2006
  8. Jun 1, 2006 #7
    I don't understand why this is so:

    [tex] \frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}[/tex]


    Isn't this correct?: [tex] cos(4y) \ = \ \sqrt{cos(4y)}^{2}[/tex]
     
  9. Jun 1, 2006 #8

    Hootenanny

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    Sorry I didn't see you had raised it to the power two, my eyes must be getting worse!

    I think the easiest way to do this is write it in terms of sec2.

    ~H
     
  10. Jun 1, 2006 #9
    Does this work:
    [tex]\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})[/tex]

    How would you write that in terms of [tex]sec^{2}[/tex]
     
  11. Jun 1, 2006 #10
    Wouldn't you have to use the method of Partial Fractions in order to get it into some form like this in order to write it in terms of sec
    [tex] \frac{A}{cos4y} + 1[/tex]
     
  12. Jun 1, 2006 #11
    I have absolutely no idea what you mean by the easiest way would be to get it in terms of sec^2
     
  13. Jun 1, 2006 #12

    TD

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    I'm having the feeling you're making this harder than necessary.
    Remember that (tan(2y))' = 2/(cos²(2y)).

    [tex]
    \frac{{dy}}{{dx}} = \left( {\cos ^2 x} \right)\left( {\cos ^2 \left( {2y} \right)} \right) \Leftrightarrow \frac{{dy}}{{\cos ^2 \left( {2y} \right)}} = \cos ^2 xdx \Leftrightarrow \frac{1}{2}\tan \left( {2y} \right) = \frac{1}{4}\left( {\sin \left( {2x} \right) + 2x} \right) + C
    [/tex]
     
  14. Jun 1, 2006 #13

    Hootenanny

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    Yes, using partial fractions;

    [tex]\frac{1}{\cos(4y) +1} = \frac{1}{2}\sec^{2}(2y)[/tex]

    Now, you sould find;

    [tex]\int \frac{1}{2}\sec^{2}(2y) \; dy[/tex]

    A little easier to solve.

    ~H
     
  15. Jun 1, 2006 #14

    TD

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    But we started with sec²(2y), that was the factor cos²(2y) at the rhs...

    Perhaps I'm missing something but I think you just went in a circle, from cos²(2y) to the double angle (losing the square) and now back introducing the square, with the half angle!
     
  16. Jun 1, 2006 #15
    Wow. Thanks ALOT! I really do have a tendancy to make things more difficult than they really are. Hey TD, could you check out the last question in the thread I started yesterday. Thanks.
     
  17. Jun 1, 2006 #16

    Hootenanny

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    Good point, guess my maths is a bit rough, best stick to physics...

    ~H
     
  18. Jun 1, 2006 #17

    TD

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    We're all here to help (or to get help), I'd probably make similar mistakes when attempting to answer physics questions :blushing:

    It's nice to see you're helping arround a lot, being a science advisor and homework helper. I remember you from a couple of months back, with a few calculus questions :smile:
     
  19. Jun 1, 2006 #18

    Hootenanny

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    Hmm, so do I, volumes of revolution and integration by substitution (one where I actually forgot to integrate) once again, my careless mistakes let me down:blushing: . Thanks again for helping me :smile:

    ~H
     
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